Question

If $P$ be any point on the plane $lx+my+nz=p$ and $Q$ be a point on the line $OP$ such that $OP.OQ=p^2$, then the locus of the point $Q$ is:

• A. $p(lx-my-nz)=x^2+y^2+z^2$
• B. $p(lx+my+nz)=x+y+z$
• C. $p(lx+my+nz)=x^3+y^3+z^3$
• D. $p(lx+my+nz)=x^2+y^2+z^2$

Answer 1

The correct option is D $p(lx+my+nz)=x^2+y^2+z^2$

Let $P(\alpha ,\beta ,\gamma )$ and $Q(x_1,y_1,z_1)$ be the two points.
$D.R^{\prime }s$ of $OP$ are $(\alpha ,\beta ,\gamma )$ and that of $OQ$ are $(x_1,y_1,z_1)$
$\because O,Q,P$ are collinear.
$\therefore \frac{\alpha }{x_1}=\frac{\beta }{y_1}=\frac{\gamma }{z_1}=k\left(say\right)\cdots \left(i\right)$
Since, $P(\alpha ,\beta ,\gamma )$ lie on the plane $lx+my+nz=p$, we have $l\alpha +m\beta +n\gamma =p$
Now using equation $\left(i\right)$
$\Rightarrow kl{x}_1+km{y}_1+kn{z}_1=p\cdots \left(ii\right)$
Since $OP.OQ=p^2$
$\therefore \sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2}\sqrt{x_1^2+y_1^2+z_1^2}=p^2$
$\Rightarrow \sqrt{k^2{x}_1^2+k^2{y}_1^2+k^2{z}_1^2}\sqrt{x_1^2+y_1^2+z_1^2}=p^2$
$\Rightarrow k(x_1^2+y_1^2+z_1^2)=p^2\cdots \left(iii\right)$
From equation $\left(ii\right)$ and $\left(iii\right)$ we get:
$\frac{l{x}_1+m{y}_1+n{z}_1}{x_1^2+y_1^2+z_1^2}=\frac{1}{p}$
Hence, locus of $Q$ is $p(lx+my+nz)=x^2+y^2+z^2$